1A). A parallel plate capacitor with an air gap has a capacitance of 70F when th

Question

1A). A parallel plate capacitor with an air gap has a capacitance of 70F when the plates are 1.4mm apart. If it stores 0.1mC on each of the plates, what is the potential difference between the plates? What is the electric field between the plates?

B). For the capacitor in the previous problem, the air gap is replaced with water (k = 80), but the charge on the plates is kept the same. What is the electric potential difference now? What is the electric field now?

C). The capacitor in part one again has its air gap replaced with water, but now the electric potential is held constant instead of the charge. What is the new amount of charge stored on the plates of the capacitor? What is the electric field between the plates?

2). A capacitor is connected to a 14 V source of voltage. If it is able to accumulate a charge of 19C on each plate, what is the capacitance of the capacitor? What will the capacitance be if the air gap is replaced with water?

Explanation / Answer

1A) C = eo*A/d = Q/V

potential difference is V = Q/C = 0.1*10^-3/(70*10^-6) = 1.42 V

electric field is E = V/d = 1.42/(1.4*10^-3) = 1014.3 V/m

B) Cnew = k*C= 80*70*10^-6= 5600*10^-6 F

Potential difference is V = Q/C = 0.1*10^-3/(5600*10^-6) = 0.017857 V

Electric field is E = V/d = 0.01785/(1.4*10^-3) = 12.75 V/m

C) Qnew = C*V = 5600*10^-6*1.42 = 0.007952 C = 7.95 mC

Electric field is E = V/d = 1.42/(1.4*10^-3) = 1014.3 V/m

2) C = Q/V = 19*10^-6/14 = 1.35 uF

Cnew = 80*C = 80*1.35 uF = 108 uF

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